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The friction force is a type of force that resists the sliding or rolling motion when the surface of one object comes in contact with the surface of another.
- It is divided into four types, namely static friction, fluid friction, sliding friction and rolling friction.
- When a body attempts to slide over another body, force opposing the object arises as a response to the applied force and acts in the opposite direction.
- Static friction exists between two bodies in contact with each other.
- ‘Limiting friction ’ is a term used to describe the limiting value of static friction.
- The most common example of frictional force is walking on a slippery surface.
- When we walk on a slippery surface, the friction force is less, and we tend to slip on these surfaces.
This experiment aims to study the relationship between the force of limiting friction and normal reaction and fill the coefficient of friction between a block and a horizontal surface.
Key Terms: Frictional Force, Force, Friction, Static Friction, Limiting Friction, Energy, Coefficient of Friction, Fluid Friction
Aim
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The study aims to identify the relationship between the force of limiting friction and the normal reaction and to find the coefficient of friction between a block and a horizontal surface.
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Material Required
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The material required for the experiment to study the relationship between force of limiting friction and normal reaction are as follows:
- Wooden Block
- 50g or 20g Weights
- Frictionless pulley at one end of a horizontal plane
- Pan
- Spring Balance
- Thread
- Spirit Level
Theory
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Sliding friction is the friction that occurs between two surfaces when bodies slide across them. It is popular with the name Kinetic friction. The sliding friction force refers to the resistance offered by two objects when sliding against each other.
Example of Sliding FrictionHere are some examples of sliding friction:
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What is force of sliding friction?
The force of sliding friction is the minimum amount of force required to make a body slide over the surface. It is mathematically represented as:
F ∝ RF
F= μR
Where
- μ is the coefficient of friction
- R is the normal reaction
- F is the force of sliding
- Net force at equilibrium should be zero.
F = P + p
R = W + w
Diagram
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The diagram to determine the relationship between the force of limiting friction and normal reaction is as follows:
Relationship between the force of limiting friction and normal reaction
Procedure of Experiment
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The procedure of the experiment are as follows:
- Ensure the pulley is free of friction and the tabletop is clean.
- It would help if you weighed the block before placing it on the tabletop.
- Tie one end of the thread to the hook of the wooden block and pass it over the block to the pulley.
- Determine the pan's weight.
- The pan should be hung vertically and fastened to the free end of the thread.
- Put some weights on the pan to make it easier to lift the block.
- Begin tapping the tabletop to make the block slide.
- Continue tapping while gradually increasing the weights.
- In the observation table, record the total weights in the pan.
- Add one 50g or 20g weight to the wooden block and repeat steps 8 and 9.
- Repeat steps 8, 9, and 10 six times each time the weight increases by 50g or 20.
- Lastly, record each of the observations in the observation table.
Observations
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The observation for the experiment are as follows:
- Weight of wooden block, W = ………….gwt
- Weight of pan, P = …………g wt.
| Sl.no | Weights on a wooden block (in g wt) | The total weight being pulled (W+w)= Normal reaction (in g wt) | Weight on the pan (p) (in g wt) | Total weight pulling the block and weights (P+p)=limiting friction (F) (in g wt) |
|---|---|---|---|---|
| 1 | ||||
| 2 | ||||
| 3 | ||||
| 4 |
Calculations
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The force of sliding friction is determined by the total weight dragging the block and weights. Total weights being pulled on a horizontal surface produce typical response R. The dynamic friction F is determined by the total weight pulling these weights.
- Plot a graph with R on the x-axis and F on the y-axis between normal response R and limiting friction F.
- As demonstrated below, the graph becomes a straight line:
Graph Between R and F
Result
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The force of limiting friction grows in proportion to the entire weight drawn. There is a direct correlation between the rise and the increase in the total weight and force of limiting friction.
- The normal response R is proportional to the limiting friction F, as seen in the graph.
- It's a pact with the friction-limitation law. (This experiment might be used to prove the law.)
- The coefficient of friction is the constant ratio FIR (p).
- The slope of the graph may be used to compute it.
Precautions and Sources of Error
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The following precautions should be taken during the experiment:
- The table should have a horizontal top.
- Between the block and the pulley, the thread should be horizontal.
- The weight should be applied in small amounts, and the pan should not fluctuate or revolve.
- Taping the tables should be done with care.
- It's best to utilize a frictionless pulley.
Sources of Error
The sources of error are as follows:
- The tabletop is not perpendicular to the ground.
- The thread connecting the block and the pulley may not be horizontal.
- The frictionless pulley must not be used.
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Things to Remember
- Friction is a contact force similar to the normal force that resists when two things slide against each other.
- The normal force works perpendicular to the flat surface of an item, whereas friction acts in the flat surface direction.
- It is observed that the total weight drawn and the limiting friction force both rise at the same time.
- The thread between the block and the pulley is kept horizontal to ensure that the whole weight of the pan and weights is distributed evenly.
- When it rains, the roads become slippery because there is less friction between the feet and the road due to the thin layer of wetness between them.
- The brake's surface is flat because hydraulic pistons push it, and the brake rotor's surface is when it is compressed.
Sample Questions
Ques. During peddling of a bicycle, the force of friction exerted by the ground on the two wheels is such that it acts in the backward direction on the front wheel and in the forward direction on the rear wheel. Why? (2 marks)
Ans. The rear wheel's point of contact tends to slide backwards when pedalling. As a result, the frictional force operates in the forward direction, opposing the rearward inclination. The front wheel, on the other hand, is propelled forward by the back wheel. The front-wheel acts backwards to counteract this frictional force.
Ques. A force of 200 N is exerted on a snack box of 5 kg still on the floor. If the coefficient of friction is 0.3, calculate the static friction? (2 marks)
Ans. Fn (Normal force) = 200 N,
μs (Coefficient of friction) = 0.3,
Static friction is given by Fs =μs Fn
= 0.3 × 200 N
Fs = 60 N.
Ques. A horizontal force of 10 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. What is the weight of the block? (2 marks)
Ans. Friction balances the weight of the body
Frictional force f = μN = mg
f= 0.2 X 10 = 2N
Therefore, weight of the block mg= 2N
Ques. A marble block of mass 2 kg lying on ice when given a velocity of 6 m/s is stopped by friction in 10s. Then what is the coefficient of friction? (consider g =10m/s) (3 marks)
Ans. u = 6 m/s
v = 0
t =10s
a = -f/m = – μmg/m = – μg = -10μ
Substituting values in v = u+at
0 = 6 – 10μ x 10
Therefore, μ = 0.06
Ques. The upper half of an inclined plane with inclination Φ is perfectly smooth, while the lower half is rough. A body starting from rest at the top will again come to rest at the bottom. Then what is the coefficient of friction for the lower half? (3 marks)
Ans. Assume that the plane is L metres long. The rise in K.E will be equal to the reduction in P.E as the block slides down the plane.
Work done, W = Change in K.E (ΔK) = (1/2) mu2 – (1/2)mv2= 0
Work done by friction (Wf) + Work done by gravity (Wg) = 0
-μgcosΦ(L/2) + mgLsinΦ = 0
⇒ (μ/2) cosΦ = sinΦ
⇒ μ = 2tanΦ
Ques. Consider a car moving on a straight road with a speed of 100 m/s. What is the distance at which a car can be stopped? (μk = 0.5) (3 marks)
Ans. When a car stops because of friction, its retarding force is
ma = μR
ma = μmg
a= μg
Consider the equation v^2 = u^2 -2as (the car is retarding so a is negative)
0 = u^2 -2as
2as = u^2
s = u^2/2a
s = u^2/2μg
s = (100)^2/2 x 0.5 x 10
s = 1000 m
Ques. The minimum force required to start pushing a body up a rough (frictional coefficient μ) inclined plane F1 while the minimum force needed to prevent it from sliding down is F2. If the inclined plane makes an angle θ from the horizontal such that tan θ= 2μ then what is the ratio F1/F2? (3 marks)
Ans. To push upwards, we must work against the sliding force due to gravity and friction. Therefore, the force F1= mgsinθ + μmgcosθ.
To stop the body from sliding work is done against sliding force but frictional force stops the body from sliding.
Therefore, the force F2= mgsinθ – μmgcosθ.
F1/F2 = (mgsinθ + μmgcosθ)/ (mgsinθ – μmgcosθ)
F1/F2 = (tanθ +μ) / (tanθ – μ)
(since tanθ = 2μ)
F1/F2 = (2μ +μ) /(2μ – μ)
F1/F2 = 3
Ques. Amy is hauling a toy car of mass 4 kg which was at rest earlier on the floor. If 50 N is the value of the static frictional force, calculate the friction coefficient? (3 marks)
Ans. Known:
m (Mass) = 4 kg,
Fs (static frictional force) = 50 N,
Fn (Normal force) = mg
= 4 Kg × 9.8 m/s2
Fn= 39 N
μs = Fs /Fn
μs = 50/39
μs = 1.282
Ques. A horizontal force of 20 N is necessary to just hold a block stationary against a wall. The coefficient of friction between the block and the wall is 0.2. What is the weight of the block? (2 marks)
Ans. Friction balances the weight of the body
Frictional force f = μN = mg
f= 0.2 X 20 = 4N
Therefore, weight of the block mg= 4 N
Ques. Amy is hauling a toy car of mass 8 kg which was at rest earlier on the floor. If 100 N is the value of the static frictional force, calculate the friction coefficient? (3 marks)
Ans. Known:
m (Mass) = 8 kg,
Fs (static frictional force) = 100 N,
Fn (Normal force) = mg
= 8 Kg × 9.8 m/s2
Fn= 78 N
μs = Fs /Fn
μs = 100/78
μs = 1.282
Ques. A marble block of mass 20 kg lying on ice when given a velocity of 5 m/s is stopped by friction in 10s. Then what is the coefficient of friction? (consider g =10m/s) (3 marks)
Ans. u = 5 m/s
v = 0
t =10s
a = -f/m = – μmg/m = – μg = -10μ
Substituting values in v = u+at
0 = 5 – 10μ x 10
Therefore, μ = 0.05
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| Class 11 Physics Study Guides | ||
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| NCERT Solutions for Class 11 Physics | Topics for Comparison in Physics | Formulas in Physics |
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